THE HADWIGER FINSLER INEQUALITY REVERSE IN AN ACUTE TRIANGLE sorin radulescu, marian dinca, marius dragan, eduard puschin

THE HADWIGER-FINSLER INEQUALITY REVERSE IN AN
ACUTE TRIANGLE
Prof.dr. Sorin Rădulescu,Prof. Marian Dincă,Prof.Marius Drăgan,Prof.
Eduard Puschin
In the paper1 the authors proposed the following inequality:
In any acute triangle ABC are true the following inequality:
a 2 + b 2 + c 2 ≤ 4 3 F + 2− 3 a − b 2 + b − c 2 + c − a 2

1

3−2 2

Abstract: In this paper for begining we shall prove the inequality 1 for the isosceles acute
triangle
and then for any acute triangle.
Theorem 1.In any isoscel acute triangle ABC with the sides of lengths a, b, c
and area F and b = c are true the inequality 1
Proof :The inequality 1 may be written as :
a 2 +b 2 +c 2 −4 3 F
≤ 2− 3
2
a−b 2 +b−c 2 +c−a 2
3−2 2

Using the notation b = c = x, because F =
inequality 2
may be written as
a 2 +b 2 +c 2 −4 3 F
a−b 2 +b−c 2 +c−a 2

;

4x 2 sin 2

=

+2x 2 −2 3 x 2 sinA
2

A
2

−x

4−4 cosA⋅ 12 +

3
2

2 2xsin

x 2 sinA and a = 2xsin A2 , the right side ratio of

4sin 2

=

A
2

+2−2 3 sinA

=

2

2 2sin A2 −1

21−cosA+2−2 3 sinA
2 2sin A2 −1

2

3
4−2 cosA+ 3 sinA

=
=

A
2

1
2

2 2sin A2 −1
π
cos 2 A4 − 12
A
4

cos 2

π
+ 12

2

=
1+cos

=
1+cos

⋅sinA

=

2

A
2
A
2

2 2sin A2 −1
− π6
1+cos

=

+ π6

1+cos

We shall consider the function f : 0,

A
2
A
2

π

2
π
6

8 sin A2 − 12
− π6
+ π6

8sin 2

4 1−cos A− π3

−1 +1 =

→ R,

fA =

A
2

=

2

8 sin A2 −sin π6
cos A2 − π6 −cos
1+cos
2sin A2 sin π6

32sin 2

− π6

A
2

2

+ π6

A
2

=
+ π6

32sin 2

A
4
A
4

π
− 12
cos 2
π
− 12
cos 2

A
4
A
4

+1

+1

1+cos A2 + π6 
5π
< π2 , or
12

= 3π
+ 2π

=
in on equivalent form
As A ≤ π2 , we have A2 + π6 ≤ π4 +
12
12
A
π
π
π
A
π
cos 2 + 6 ≥cos 4 + 6 also sin 2 ≤sin 4
it follow that fA ≤ f π2  and we have the inequality of the statement.
Theorem 2. For any triangle ABC of semiperimeter s with incircle CI, r and circumcicle
CO, R there exist two isosceles triangles
A 1 B 1 C 1 and A 2 B 2 C 2 of semiperimeter s 1 and s 2 with incircle CI, r and circumcicle CO, R
So as inequality exist :
s2 ≤ s ≤ s1
4
Proof. The straight line OI cat the circumcircle in it A 1 and A 2 .The tangents to the incircle
from a point A 1 , cat the circumcicle in
B 1 and C 1 .The straight line B 1 C 1 is tangent to the incircle ,acording to Poncelet’s closure
theorem.
The triangle A 1 B 1 C 1 is isosceles. Similarly the tangents to the incircle from a point A 2 cat
the circumcircle in B 2 and C 2 .
The straght line B 2 C 2 is tangent to the incircle ,acording to Poncelet’s closure theorem it
follows that A 2 B 2 C 2 is isoscel.
As a 1 = B 1 C 1 = 2rR+r+d
, b 1 = c 1 = A 1 C 1 = A 1 B 1 = R+r+dR+d
, where

2 2
2 2
R+d −r

R+d −r

2

d =∣ OI ∣= R − 2Rr and
a 2 = B 2 C 2 = 2rR+r−d
, b2 = c2 = A2C2 = A2B2 =
2 2
R−d −r

s1 =

1
2

a 1 + b 1 + c 1  =

R+r−dR−d
R−d 2 −r 2

and as

2R 2 + 10Rr − r 2 + 2R − 2r R 2 − 2Rr

π
− 12

π
+ 12


s 2 = 12 a 2 + b 2 + c 2  = 2R 2 + 10Rr − r 2 − 2R − 2r R 2 − 2Rr
acording with the W.J.Blundon inequality it follows that the inequality 4 is true
Proof of the inequality (1) If the point O exterior of the incircle C(I,r),rezult∣ OI∣= d ≥ r
, and the tangent to the incircle from a point O,cat
the circumcircle in B 3 and C 3 and the tangent to the incircle C(I,r) from a points B 3 and
C 3 cat the circumcircle C(O,R) in A 3 ,
acording the Poncelet’s closure theorem.The triangle A 3 B 3 C 3 it is right angled
∢B 3 A 3 C 3 = π2 ,
Let s the semiperimeter of the acute angled triangle ABC and s 3 the semiperimeter of the
triangle A 3 B 3 C 3
We shall prove that s ≥ s 3
5
Let the triangle A 3 B 3 C 3 tangent to the incircle C(I,r) in the points ; D.E.F ;where
D ∈ B 3 C 3 ; E ∈ C 3 A 3 and F ∈ B 3 A 3
we have A 3 F = A 3 E = x = rctg π4 = r, B 3 F = B 3 D = y = rctg B23 ,
C 3 D = C 3 E = z = rctg C23 ,
As B 3 C 3 = B 3 D + C 3 D = y + z = 2R ,we shall obtain :s 3 =x + y + z = 2R + r ,it rezults that
the inequality 5 is equivalent to :
R ∑ sinA ≥ r + 2R ,or ∑ sinA ≥ Rr + 2 = ∑ cosA + 1, or ∑ sinA −cosA ≥ 1 ,6 lets
ciclic
π−α
,B
2

ciclic

ciclic

ciclic

= π−β
, C = π−γ
, as A, B, C ∈ 0, π2 
2
2
we have that α, β, γ ∈ 0, π ,and α + β + γ = π .The inequality 6 is equivalented
to ∑ cos α2 −sin α2  ≥ 1,or ∑ cos α2 −tg π4 sin α2  ≥ 1, or

A=

ciclic

ciclic

∑ cos α2 +

ciclic

π
4

 ≥cos π4

7, let

α

2

+

π
4

= u,

β
2

+

π
4

= v and

γ
2

+

π
4

= w and

using well-know identity:

cosu +cosv +cosw +cosu + v + w = 4 ∏ cos u+v
 it follows that:
2
ciclic

∑ cos α2 +

ciclic

π
4

 +cos ∑  α2 +

or ∑ cos α2 +
ciclic

π
4



ciclic
−cos π4

π
4

 = 4 ∏ cos α+β
+

4
ciclic

= 4 ∏ cos π2 −
ciclic

γ
4

π
4

 = 4 ∏ cos π−γ
+
4
ciclic

π
4



 = 4 ∏ sin γ4 ≥ 0 we shall obtain the
ciclic

inequality 7
Using well-known identity: ab + bc + ca = s 2 + r 2 + 4Rr and
a 2 + b 2 + c 2 = 2s 2 − 2r 2 − 8Rr and F = sr, let 2− 3 = λ
3−2 2

The inequality 1 is equivalented to
4 3 sr + 2λ − 12s 2 − 2r 2 − 8Rr − 2λs 2 + r 2 + 4Rr ≥ 0
or λ − 1s 2 + 2 3 sr + 1 − 3λr 2 + 4 − 12λRr = Es, R, r ≥ 0
8
As λ > 1, we shall obtain : Es, R, r ≥ Es 3 , R, r
9
In order to prove inequality 8 it will be sufficient to prove inequality Es 3 , R, r ≥ 0 ,what
reprezent the inequality 1 for right angled triangle
A3B3C3
bc
2
2
2
We have F = 2 , and a = b + c .We shall obtain the following inequality:
b
c
2 3 bc + 2λ − 12b 2 + c 2  − 2λ b 2 + c 2 b + c + bc ≥ 0, let b+c
= x and b+c
=y
obtain
2 3 xy + 2λ − 121 − 2xy − 2λ 1 − 2xy + xy ≥ 0 or
xy 3 + 2 − 5λ + 2λ − 1 ≥ λ 1 − 2xy ,let xy = t
As x + y = 1, we shall obtain : t ∈ 0; 14  ,let 1 − 2xy = θ ,θ ∈  1 , 1, it will rezult that
:xy

2
= 1−θ
2
1−θ 2


2

2

,we shall obtain:
3 + 2 − 5λ + 2λ − 1 − λ ⋅ θ ≥ 0

or 5λ − 3 − 2θ 2 − 2λθ − λ + 3 ≥ 0 ,

the equation 5λ − 3 − 2θ 2 − 2λθ − λ + 3 = 0 has a root θ 1 =
it follows that θ 2 =

2 −λ+ 3 
5λ− 3 −2

1
2

and θ 1 θ 2 =

−λ+ 3
5λ− 3 −2


1
2

and θ 2 <

and 5λ − 3 − 2θ 2 − 2λθ − λ + 3 = 5λ − 3 − 2θ − θ 1 θ − θ 2  ≥ 0

For ∣ OI ∣= d ≤ r, it shall rezult that any triangle is acute, let ∢B 1 A 1 C 1 = α and
∢B 2 A 2 C 2 = β
r
r
sin α2 = R+d
< 1 =sin π4 , it follows that α2 < π4 and α < π2 .As sin β2 = R−d
≤ 1 or
2

2

r 2 ≤ R − d , or d ≤ R − r 2 , or
d 2 ≤ R − r 2  2 or R 2 − 2Rr ≤ R 2 − 2 2 Rr + 2r 2 or  2 − 1R ≤ r or R ≤  2 + 1r
,this inequality is true,because
d ≤ r imply R 2 − 2Rr ≤ r 2 , or R − r 2 ≤ 2r 2 who imply R ≤  2 + 1r we shall obtain:
β
sin 2 ≤ 1 =sin π4 , it follows that β ≤ π2
2

Because Es, R, r ≥ Es 2 , R, r and Es 2 , R, r ≥ 0 because is an inequality for acute
triangle and isosceles triangle
We have inequality of the statement
References
1 Cezar Lupu,Constatin Mateescu,Vlad Matei,Mihai Opincariu:Refinements of the
Finsler-Hadwiger
reverse inequality,Gazeta Matematica seria A,nr 1-2(2010)p.49-53
2 P.G.Popescu,I.V.Maftei,J.L.Diaz-Barrero,M.Dincă:Inegalitati Matematice;Modele
Inovatoare,
Editura Didactica şi Pedagogica,2007

3 Marian Dincă,J.L.Diaz-Barrero: A new proof of an Inequality of Oppenheim, vixra
org:1008.0013
4 Marian Dincă,Mihaly Bencze:New generalisation Finsler-Hadwiger inequality:Octogon
Mathematical Magazine,2002